A law enforcement officer in an intergalactic "police car" turns on a red flashing light and sees it generate a flash every 1.5 s. A person on earth measures that the time between flashes is 2.9 s. How fast is the "police car" moving relative to the earth?
Solve this equation for v:
2.9 = 1.5 / sqrt [1 - (v/c)]^2
c is the speed of light.
This is one of Einstein's relativity equations
To solve this problem, we can use the concept of time dilation from special relativity. Time dilation occurs when an object is moving relative to an observer, causing time to appear to run slower for the moving object compared to the stationary observer.
Let's assume that the police car is moving relative to the Earth. We need to find the velocity of the police car (v) using the time dilation formula:
Δt' = γΔt
Where:
Δt' is the time observed on Earth (2.9 s)
Δt is the proper time (1.5 s)
γ is the Lorentz factor, given by γ = 1 / √(1 - (v^2 / c^2))
In this formula, c represents the speed of light, which is a known constant value of approximately 3 x 10^8 meters per second.
Now, let's solve for v:
1. Calculate γ:
γ = 1 / √(1 - (v^2 / c^2))
γ = 1 / √(1 - (v^2 / (3 x 10^8)^2))
2. Substitute the values of γ and Δt' into the time dilation equation:
Δt' = γΔt
2.9 s = γ * 1.5 s
3. Rearrange the equation to solve for γ:
γ = 2.9 s / 1.5 s
4. Plug in the value of γ into the Lorentz factor equation:
2.9 s / 1.5 s = 1 / √(1 - (v^2 / (3 x 10^8)^2))
5. Square both sides of the equation to solve for v:
(v^2 / (3 x 10^8)^2) = (1 - (2.9 s / 1.5 s))^2
6. Simplify the equation:
(v^2 / (3 x 10^8)^2) = (1 - (1.93333333))^2
7. Calculate the square root of both sides to isolate v:
v^2 = (1 - (1.93333333))^2 * (3 x 10^8)^2
8. Take the square root of both sides to solve for v:
v = √[(1 - (1.93333333))^2 * (3 x 10^8)^2]
After performing the calculations, you will find the value of v, which represents the velocity of the police car relative to the Earth.